Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the value of \(k\) such that the line \(k x-3 y=10\) (a) is parallel to the line \(y=2 x+4\); (b) is perpendicular to the line \(y=2 x+4\); (c) is perpendicular to the line \(2 x+3 y=6\).

Step-by-Step Solution

Answer

(a) \(k = 6\); (b) \(k = -\frac{3}{2}\); (c) \(k = \frac{9}{2}\).

Step 1: Understand Parallel Lines

Two lines are parallel if they have the same slope. The equation given is \(y = 2x + 4\), which has a slope of 2. We need to rearrange the equation \(kx - 3y = 10\) to find its slope. First, express it in the form \(y = mx + c\): \(3y = kx - 10\), then \(y = \frac{k}{3}x - \frac{10}{3}\). Therefore, the slope of this line is \(\frac{k}{3}\). To be parallel, set \(\frac{k}{3} = 2\).

Step 2: Solve for k (Parallel Case)

Set \(\frac{k}{3} = 2\) from Step 1. Solve for \(k\) by multiplying both sides by 3, resulting in \(k = 6\).

Step 3: Understand Perpendicular Lines

Two lines are perpendicular if the product of their slopes is -1. The slope of the line \(y = 2x + 4\) is 2. Thus, the slope of a line perpendicular to it is \(-\frac{1}{2}\). Again, express \(kx - 3y = 10\) in slope-intercept form as \(y = \frac{k}{3}x - \frac{10}{3}\). The slope of this line is \(\frac{k}{3}\). So, set \(\frac{k}{3} \times 2 = -1\).

Step 4: Solve for k (Perpendicular to Given Line)

Set \(\frac{k}{3} \times 2 = -1\). Solve for \(k\): \(\frac{2k}{3} = -1\). Multiply by 3 to get: \(2k = -3\). Then, divide by 2: \(k = -\frac{3}{2}\).

Step 5: Rearrange Second Given Equation

Rearrange the equation \(2x + 3y = 6\) into the slope-intercept form: \(3y = -2x + 6\), thus \(y = -\frac{2}{3}x + 2\). The slope is \(-\frac{2}{3}\). For a line to be perpendicular to this, its slope must be the negative reciprocal, which is \(\frac{3}{2}\).

Step 6: Solve for k (Perpendicular to Second Given Line)

Set the slope \(\frac{k}{3}\) equal to \(\frac{3}{2}\) since this is the negative reciprocal of \(-\frac{2}{3}\). Setting \(\frac{k}{3} = \frac{3}{2}\), solve for \(k\) by multiplying both sides by 3, yielding \(k = \frac{9}{2}\).

Key Concepts

Slope-Intercept Form

The slope-intercept form is a straightforward way to express the equation of a line. In this form, a line is represented as \( y = mx + b \), where:

• \( m \) stands for the slope of the line, showing how steep the line is.
• \( b \) is the y-intercept, indicating where the line crosses the y-axis.

To convert an equation into slope-intercept form, rearrange it to solve for \( y \). For example, given an equation \( kx - 3y = 10 \), you can rearrange it to \( y = \frac{k}{3}x - \frac{10}{3} \).
This rearrangement helps you identify key components like slope, which is necessary to determine whether lines are parallel or perpendicular.

Equation of a Line

The equation of a line is a mathematical statement that describes a straight line's graph. It reveals important characteristics such as slope and y-intercept. - There are various forms, but the slope-intercept form \( y = mx + b \) is commonly used for comparing lines.- For instance, an equation like \( y = 2x + 4 \) directly shows a slope of 2 and a y-intercept of 4.Converting equations like \( kx - 3y = 10 \) into this form clarifies comparisons by highlighting the slope. This is crucial when determining relationships between lines, such as parallelism or perpendicularity.

Negative Reciprocal

The negative reciprocal concept is essential for understanding perpendicular lines. Two lines are perpendicular if the product of their slopes is \(-1\). In simpler terms, the slope of one line is the negative reciprocal of the slope of the other.- For example, if one line has a slope of 2, a perpendicular line would have a slope of \(-\frac{1}{2}\), calculated by flipping the original slope and changing the sign.- When rearranging \( 2x + 3y = 6 \) to \( y = -\frac{2}{3}x + 2 \), the slope becomes \(-\frac{2}{3}\). A line perpendicular to this would then have a slope of \( \frac{3}{2} \).This understanding helps solve the problem of finding the value of \( k \) when the line needs to be perpendicular to another.

Solving Linear Equations

Solving linear equations involves finding the value of variables that satisfy the equation. It's a foundational concept in algebra and helps solve real-world problems by representing relationships between quantities. - For the equation \( \ kx - 3y = 10 \), our goal is to find \( k \) such that it fulfills certain conditions, like being parallel or perpendicular to another line.- Understanding how to manipulate the equation into slope-intercept form is crucial for determining the value of \( k \) that meets the problem's criteria.This process of rearrangement and comparison allows us to solve the original problem step-by-step, ensuring we account for all necessary conditions for parallel or perpendicular relationships.